For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging exponential profile starting from 1, which perfectly matches the curve in option (4). The expression eKt dictates an exponential increase. Ensure the curve starts from a non-zero value (1) at t=0, as (N0/N0) = 1, rather than starting from the origin (0). Class 12 Chemistry: Chemical Kinetics

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging exponential profile starting from 1, which perfectly matches the curve in option (4). The expression eKt dictates an exponential increase. Ensure the curve starts from a non-zero value (1) at t=0, as (N0/N0) = 1, rather than starting from the origin (0). Class 12 Chemistry: Chemical Kinetics

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the…

Year	2026	2025	2024	Total
Questions	14	20	8	42

Q66 jee_main_2026_24_january_morning Arrhenius Equation and Catalyst Effect

At

27°C

in presence of a catalyst, activation energy of a reaction is lowered by

10 kJ mol⁻¹

. The logarithm ratio of

k(catalysed)k(uncatalysed)

is (Consider that the frequency factor for both the reactions is same)

A. $17.41$ $17.41$
B. $1.741$ $1.741$
C. $3.482$ $3.482$
D. $0.1741$ $0.1741$

Solution

Related Formula

k = A e-(Eₐ)/(RT)

Core Logic

Let the activation energy of the uncatalyzed reaction be $Eₐ$ $E_a$. Activation energy of the catalyzed reaction $= Eₐ - 10 kJ mol⁻¹$ $= E_a - 10 \text{ kJ mol}^{-1}$. $kuncatalyzed = A e-(Eₐ)/(RT)$ $kcatalyzed = A e-((Eₐ - 10000))/(RT)$

Dividing the two rate constants: $kcatalyzedkuncatalyzed = e(Δ Eₐ)/(RT)$ where $Δ Eₐ = 10 kJ mol⁻¹ = 10000 J mol⁻¹$ .

Step 1: Calculate Logarithm Ratio

Taking natural logarithm on both sides:

ln ( kcatalyzedkuncatalyzed) = (Δ Eₐ)/(RT)

Converting to base 10 logarithm:

₁₀ ( kcatalyzedkuncatalyzed) = (Δ Eₐ)/(2.303 RT)

Substitute the values ( $R = 8.314 J K⁻¹ mol⁻¹$ , $T = 300 K$ $T = 300 \text{ K}$):

₁₀ ( kcatalyzedkuncatalyzed) = (10000)/(2.303 \times 8.314 \times 300)

₁₀ ( kcatalyzedkuncatalyzed) = (10000)/(5744.14) \approx 1.741

Pattern Recognition

Remember that lowering activation energy by $Δ Eₐ$ $\Delta E_a$ increases the rate by a factor of $eΔ Eₐ / RT$ $e^{\Delta E_a / RT}$. Taking log base 10 simply scales this exponent by $1/2.303$ $1/2.303$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q72 jee_main_2026_24_january_evening First Order Reactions

The half-life of

⁶⁵

Zn is 245 days. After x days, 75% of original activity remained. The value of x in days is ____. (Nearest integer) (Given : log 3 = 0.4771 and log 2 = 0.3010)

Numerical Answer. Answer: 102 to 102

Solution

Related Formula

t = (2.303)/(K) ((a₀)/(aₜ))

where $K = ln 2t1/2$ $K = \frac{\ln 2}{t_{1/2}}$

Core Logic

Given $t1/2 = 245 days$ $t_{1/2} = 245 \text{ days}$, so decay constant $K = (ln 2)/(245)$ $K = \frac{\ln 2}{245}$. If 75% activity remained, the fraction remaining is $aₜ / a₀ = 3/4$ $a_t / a_0 = 3/4$ (or $75/100$ $75/100$). So, $(a₀)/(aₜ) = (4)/(3)$ $\frac{a_0}{a_t} = \frac{4}{3}$.

Step 1: Substitute Values

t = (1)/(K) ln((4)/(3))

t = (245)/(ln 2) ln((4)/(3))

Converting to base 10 logs:

t = 245 \times ( (4/3))/( 2)

t = 245 [ (2 2 - 3)/( 2) ]

Step 2: Arithmetic Evaluation

t = 245 [ (2(0.3010) - 0.4771)/(0.3010) ]

t = 245 [ (0.6020 - 0.4771)/(0.3010) ]

t = 245 [ (0.1249)/(0.3010) ] = 245 \times 0.4149

t 101.66 days

Rounded to nearest integer = 102.

Pattern Recognition

All radioactive decays follow first-order kinetics. Don't waste time deriving formulas—plug $aₜ = 0.75 a₀$ $a_t = 0.75 a_0$ straight into the standard integrated rate law.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics Class 12 Physics: Nuclei

Q57 jee_main_2026_28_january_morning First Order Reaction Decomposition

An organic compound undergoes first order decomposition. The time taken for decomposition to

((1)/(8))th

and

((1)/(10))th

of its initial concentration are

t1/8

and

t1/10

respectively. What is the value of

t1/8t1/10\times 10

? (

2 = 0.3

)

A. $9$ $9$
B. $0.9$ $0.9$
C. $3$ $3$
D. $30$ $30$

Solution

Related Formula

t = (1)/(k)ln((A₀)/(Aₜ))

Step 1: Formulate Time Ratios

For $Aₜ = A₀/8$ $A_t = A_0/8$:

t1/8 = (1)/(k)ln((A₀)/(A₀/8)) = (1)/(k)ln(8)

For $Aₜ = A₀/10$ $A_t = A_0/10$:

t1/10 = (1)/(k)ln((A₀)/(A₀/10)) = (1)/(k)ln(10)

Step 2: Find the Ratio

t1/8t1/10 = (ln 8)/(ln 10) = ( 8)/( 10)

= (3 2)/(1) = 3(0.3) = 0.9

Therefore:

t1/8t1/10 \times 10 = 0.9 \times 10 = 9

Pattern Recognition

For first order kinetics, ratio of completion times is independent of $k$ $k$ and strictly equals the ratio of logs of initial-to-final concentration fractions.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q72 jee_main_2026_28_january_evening Arrhenius Equation And Activation Energy

A arrow B

(first reaction)

C arrow D

(second reaction) Consider the above two first-order reactions. The rate constant for first reaction at

500 K

is double of the same at

300 K

. At

500 K

,

50%

of the reaction becomes complete in

2 hour

. The activation energy of the second reaction is half of that of first reaction. If the rate constant at

500 K

of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at

300 K

is ____

\times 10⁻¹ hour⁻¹

(nearest integer).

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

ln( KT₂KT₁) = (Eₐ)/(R) [ (1)/(T₁) - (1)/(T₂) ]

K = ln 2t1/2 (First Order)

Core Logic

For Reaction 1 ( $A K₁ B$ $A \xrightarrow{K_1} B$): $(K₁)500K = 2(K₁)300K$ . $ln(2) = Eₐ₁R [ (1)/(300) - (1)/(500) ] = Eₐ₁R [ (200)/(150000) ] = Eₐ₁R [ (2)/(1500) ]$ $Eₐ₁ = (ln 2 \times R \times 1500)/(2)$

Also for Reaction 1 at 500K, half-life is 2 hours, so: $(K₁)500K = (ln 2)/(2) hr⁻¹$

For Reaction 2 ( $C K₂ D$ $C \xrightarrow{K_2} D$): Given $Eₐ₂ = Eₐ₁2 = (ln 2 \times R \times 1500)/(4)$ Given $(K₂)500K = 2 \times (K₁)500K = 2 \times (ln 2)/(2) = ln 2$

Applying Arrhenius equation for Reaction 2 from 300K to 500K: $ln [ (K₂)500K(K₂)300K ] = Eₐ₂R [ (1)/(300) - (1)/(500) ]$ $ln [ ln 2(K₂)300K ] = (((ln 2 \times R \times 1500)/(4)))/(R) \times (2)/(1500) = (ln 2)/(2) = ln(\sqrt(2))$

Step 1: Solve for K2 at 300K

Equating the arguments of the natural logs: $ln 2(K₂)300K = \sqrt(2)$ $(K₂)300K = ln 2\sqrt(2) = (0.693)/(1.414) = 0.49 hr⁻¹$ $(K₂)300K = 4.9 \times 10⁻¹ hr⁻¹ \approx 5 \times 10⁻¹ hr⁻¹$ (rounded to nearest integer).

Pattern Recognition

Sequential tracking of unknowns through Arrhenius. $Eₐ$ $E_a$ ratios dictate log ratios between temperatures directly.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q39 jee_main_2025_02_april_evening Reaction Mechanism and Energy Profiles

Reactant A converts to product D through the given mechanism (with the net evolution of heat):

A arrow B

slow;

Δ H = +ve

B arrow C

fast;

Δ H = -ve

C arrow D

fast;

Δ H = -ve

Which of the following represents the above reaction mechanism?

A. $Graph (1)$ $\text{Graph (1)}$
B. $Graph (2)$ $\text{Graph (2)}$
C. $Graph (3)$ $\text{Graph (3)}$
D. $Graph (4)$ $\text{Graph (4)}$

Solution

Related Formula

k = A e^-Eₐ/RT

Rate \propto 1Eₐ

Core Logic

Let us break down each step of the mechanism:

Step 1: $A arrow B$ $\mathrm{A} \rightarrow \mathrm{B}$ is slow.
Being the rate-determining step, it must have the highest activation energy barrier ( $Eₐ₁$ $E_{\mathrm{a1}}$).
Since $Δ H = +ve$ $\Delta H = +\mathrm{ve}$ (endothermic), the energy level of intermediate state B must be higher than the reactant state A.
Step 2: $B arrow C$ $\mathrm{B} \rightarrow \mathrm{C}$ is fast.
It has a much lower activation energy barrier ( $Eₐ₂$ $E_{\mathrm{a2}}$).
Since $Δ H = -ve$ $\Delta H = -\mathrm{ve}$ (exothermic), the energy level of intermediate C is lower than state B.
Step 3: $C arrow D$ $\mathrm{C} \rightarrow \mathrm{D}$ is fast.
It has a low activation energy barrier ( $Eₐ₃$ $E_{\mathrm{a3}}$).
Since $Δ H = -ve$ $\Delta H = -\mathrm{ve}$ (exothermic), the energy level of final state D is lower than state C.
Net Reaction: Exothermic with "net evolution of heat".
The potential energy of the final product D is lower than the initial potential energy of reactant A.

Step 1: Check Potential Energy Profile

Evaluating the transition states and relative energy levels in Graph (1):

The first peak (transition state 1) is clearly the highest ( $Eₐ₁ > Eₐ₂, Eₐ₃$ ) $\implies$ Step 1 is the slowest.
The intermediate B is higher in energy than A.
Intermediates C and product D are progressively lower in energy.
Product D has lower energy than reactant A (net exothermic).

Annotated reaction mechanism coordinate graph showing relative activation energies

This perfectly corresponds to Graph (1).

Pattern Recognition

Kinetics shortcut: Slow step = tallest peak. Exothermic step = drop in energy levels of products/intermediates. Endothermic step = climb in energy levels. Use these rules to visually scan energy profiles in under 5 seconds.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding the Matching Curve

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Logarithm Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute Values

Step 2: Arithmetic Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Formulate Time Ratios

Step 2: Find the Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for K2 at 300K

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Check Potential Energy Profile

Pattern Recognition

Chapter Mix

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